State whether true or false.
\[3\sqrt 2 \] is a rational number.

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Hint: Here we check from the definition of rational numbers if the number \[3\sqrt 2 \] satisfies the conditions of a rational number or not.

Complete step-by-step answer:
We know a number that can be written in the form \[\dfrac{p}{q}\] where \[q\] is non-zero and both the numerator and denominator are integers is called a rational number, i.e. the decimal representation of a rational number is either terminating or recurring.
Also, any number that is not a rational number is called an irrational number.
We know \[\sqrt 2 \] is an irrational number because it cannot be represented in the form \[\dfrac{p}{q}\], also if we try to find the square root of 2 in decimal from then, \[\sqrt 2 = 1.414213562....\] where the decimal representation is non-recurring, non-repeating and non-terminating.
Since, \[\sqrt 2 \] is irrational, therefore if we multiply any number other than zero or an irrational to \[\sqrt 2 \] then only it can become rational, otherwise multiplication by any other number will have no change on it.
So, \[3\sqrt 2 \] is also an irrational number.
So, the statement given in the question is FALSE.
So, option B is correct.

Note: Students many times make mistake of thinking the number which can be written in any decimal from will be a rational number but they have to keep in mind the number written after the decimal should be terminating or recurring for the number to be a rational number, because if the decimal is non-terminating you will never be able to write it in the form of fraction \[\dfrac{p}{q}\], therefore it has to be irrational. Also, not every number with square root is irrational as many numbers which are perfect squares can be solved by cancelling the square root by square power making the number a rational number.